use the function below to answer parts (a)-(c).\n\n$f(x)=x^{2}-4$\n\n(a) use the formal definition to find…

use the function below to answer parts (a)-(c).\n\n$f(x)=x^{2}-4$\n\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$.\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1,f(1))$\n(c) graph $y = f(x)$ and the tangent line at the point $(1,f(1))$ in the same coordinate system.\n\n(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=limlimits_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the\nfinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$.\n\n$f(1)=limlimits_{h\rightarrow0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$\n\nevaluate the limit expression to find $f(1)$.\n\n$f(1)=\\square$ (type an integer or a fraction.)

use the function below to answer parts (a)-(c).\n\n$f(x)=x^{2}-4$\n\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$.\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1,f(1))$\n(c) graph $y = f(x)$ and the tangent line at the point $(1,f(1))$ in the same coordinate system.\n\n(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=limlimits_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the\nfinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$.\n\n$f(1)=limlimits_{h\rightarrow0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$\n\nevaluate the limit expression to find $f(1)$.\n\n$f(1)=\\square$ (type an integer or a fraction.)

Answer

Explanation:

Step1: Expand the numerator

First, expand ((1 + h)^{2}-4) and ((1)^{2}-4). ((1 + h)^{2}-4=1 + 2h+h^{2}-4=h^{2}+2h - 3) ((1)^{2}-4=1 - 4=-3) So the numerator ([(1 + h)^{2}-4]-[(1)^{2}-4]=(h^{2}+2h - 3)-(-3)=h^{2}+2h)

Step2: Simplify the limit expression

The limit expression (f^{\prime}(1)=\lim_{h\rightarrow0}\frac{[(1 + h)^{2}-4]-[(1)^{2}-4]}{h}=\lim_{h\rightarrow0}\frac{h^{2}+2h}{h}) Since (h\neq0) (as (h\rightarrow0) but (h\neq0) in the limit process), we can cancel out the (h) terms. (\frac{h^{2}+2h}{h}=h + 2)

Step3: Evaluate the limit

Now, find (\lim_{h\rightarrow0}(h + 2)). Substitute (h = 0) into (h+2). (\lim_{h\rightarrow0}(h + 2)=0 + 2)

Answer:

(2)